OHMIO 2007 Individual Solutions
1. A triangle is isosceles but not a right triangle. If one of its interior angles has twice the measure of another of its interior angles, how many degrees in measure is the smallest of the triangle's interior angles?
Answer: 36°, or just 36
Solution: The three interior angles have measures x, 2x, and 2x, or else of measures x, x and 2x, depending on whether the two congruent angles are the larger or smaller interior angles. But the sum of the interior angles is 180°, so either x+2x+2x = 180 (in which case x = 36), or else x+x+2x = 180 (in which case x = 45 and 2x = 90, ruled out since it isn't a right triangle).
2.
Answer: 15π, or 15π sq. units
Solution: A semicircle of diameter d has area π(d/2)2 = πd2. So, the region has area π(10)2+ π(2)2+ π(5)2– π(3)2 = π(100+4+25–9) = 15π.
3. For positive integers c and d, let the notation c * d denote the number of digits in the result of multiplying together c ´ d. For example, 9 * 19 equals 3, since 9 ´ 19 = 171, where 171 is a 3-digit number.
Find 2007 * (2007 * (2007 * (2007 * 2007))).
Answer: 5
Solution: 2007 * 2007 = 7, the number of digits in the result of multiplying
2007 by 2007 (getting roughly 4000000). So, 2007 * (2007 * 2007)
= 2007 * 7 = 5, since 2007 ´ 7 = 14049, a 5-digit number. So,
2007 * (2007 * (2007 * 2007)) = 2007 * 5 = 5, since 2007 ´ 5 = 10035, a 5-digit number. Similarly, 2007 * (2007 * (2007 * (2007 * 2007))) = 5.
4. At an expensive restaurant, George's meal cost $1 more than Barbara's, before the tip or the 6% sales tax were added. George left the same amount of tip that Barbara did. His tip was 15% of his meal's cost (including sales tax in computing the tip) and her tip was 16% of her meal's cost (excluding sales tax in computing the tip). What was the cost of Barbara's meal, excluding the tax and the tip? Assume that no "rounding to the nearest cent" was required in computing all taxes and tips.
Answer: $159
Solution: If her meal's cost was $x before tax and tip, his was $(x+1). Since they left equal tips, .15(1.06(x+1)) = .16x. Upon multiplying by 100, we have 15(1.06x+1.06) = 16x, so 15.9x + 15.9 = 16x, so 15.9 = 0.1x, from which 159 = x.
5. Lance bicycles uphill at a rate 12 miles per hour less than his rate on level ground. In his last bicycle ride, half of his time was spent uphill, half on level ground. If he rode 300% more miles on level ground than than he rode uphill, how fast does he bicycle on level ground?
Answer: 16 miles per hour
Solution: Let r be his rate on level ground, r–8 his rate uphill, d his distance traveled uphill, and 4d his distance traveled on level ground, which is 300% more than d. Then from T = we have that
= , so = , so 4r–48 = r, or 3r = 48, and r = 16.
6. The base length of a triangle is roughly 10 in., known to be at most 0.2 in. too high or too low an estimate. The height of the triangle is roughly 2 in., known to be at most 0.1 in. too high or too low an estimate. Given no further information, the area of the triangle is therefore known to be roughly x many in.2, and known to be at most 0.7 in.2 too high or too low. Find x, expressed in decimal form.
Answer: 10.01
Solution: At most, the area is = 10.71. And, the area is at least = 9.31. So, the area is between 9.31 in.2 and 10.71 in.2, or 10.01±0.7. Thus x must be 10.01, since any other estimate could be off by more than 0.7 in.2.
7. When a fair coin is flipped ten times, find log2p, where p is the probability that no two consecutive flips have the same outcome.
Answer: –9
Solution: The outcome sequences HTHTHTHTHT and THTHTHTHTH are the ones described, so p = = 2–9, so log2p = log2(2–9) = –9.
8. Given that 6 sec4x – 11 sec2x + 4 = 0, find tan2x.
Answer:
Solution: Letting y = sec2x, we have 6y2–11y+4 = 0, so (2y–1)(3y–4) = 0, so y = , or y = . But sec2x ≥ 1, so sec2x = . Thus cos2x = , so sin2x = , and tan2x = = . OR, tan2x = sec2x – 1 = – 1 = .
9. Find the slope of the tangent line to y = x3x at the point where x = e.
Answer: 6e3e
Solution: Using logarithmic differentiation, ln|y| = 3x ln|x|. Upon implicit differentiation, y´ = 3 ln|x| + 3x = 3 ln|x| + 3. So,
y´ = y(3 ln|x| + 3) = x3x(3 ln|x| + 3). At x = e, the slope is then
y´ = e3e(3 ln|e| + 3) = e3e(3 + 3) = 6e3e.
Or, y = x3x = e3x ln x, so y¢ = e3x ln x(3x ln x)¢ = e3x ln x(3 ln x + 3x(1/x)), so when x = e the slope is then y¢ = e3e(3+3) = 6e3e.
10. Express in the form + + , where b,c,d are integers with b > c > d. Hint: how does A3–B3 factor?
Answer: + +
Solution: From the hint, since A3–B3 = (A–B)(A2+AB+B2), we have that
()3–()3 = (–)(()2++()2), so that
3–2 = (–)(++). So, multiplying on top and bottom by
++, we get = =
= ++.
11. For positive constants m and c, suppose that the triangle bounded by
x+y = 6c and y = 0 and y = mx has area equal to mc2. Find m.
Answer: 17
Solution: y = mx intersects x+y = 6c where x+mx = 6c, so x = . So, y = at their intersection. Thus the triangle, which has base width 6c and height , has area 6c = mc2 (from the given information). Upon dividing by mc2, we have that = 1, so m = 17. (Any c works.)
12. The repeating decimal equals for what integer n?
Answer: 8910
Solution: First solution: Recognize that
.
Second solution: Multiply by 100 to get . Subtracting one equation from the other, 22.3 = , so
n = 198693 ÷ 22.3 = 8910.
Third solution: n » 9000 so that 2007÷n » 2/9 = 0.222… . By trial and error of integer choices near n = 9000, one finds that 2007÷n yields on a calculator when n = 8910.
13.
Answer: , or sq. in.
Solution: By the Pythagorean theorem, the short legs have length . So, the rectangle has width and has height , so has area
( )( ) = = .
14. Suppose that f -1 is the inverse of a function f, and that -2 is the slope of the tangent line to y = f(x) at the point (-1,3). Find the slope of the tangent line to y = f -1(x) at the point (3,-1).
Answer: -1/2
Solution: The 2nd tangent line will be the mirror reflection of the original tangent line, through the line of symmetry y = x. So, instead of it falling 2 units from (-1,3) to (0,1) from a "run" of 1 unit, the 2nd tangent line falls 1 unit from (1,0) to (3,-1) from a "run" of 2 units, so has slope -1/2.
15. Simplify the sum log70( ) + log70( ) + log70( ) + ...+ log70( ).
Answer: -2
Solution: The sum combines into the single log
log70( ... ). Massive cancellation yields
log70( ) = log70( ) = -2.
16. Consider the sequence of functions f(x) = 1+2x, f(x) = 2+3x+4x2,
f(x) = 3+4x+5x2+6x3, f(x) = 4+5x+6x2+7x3+8x4, etc. Find f(-1), where f is the 2007’th function in the sequence.
Answer: -1004
Solution: f(x) = 2007+2008x+2009x2+2010x3+...+4013x2006+4014x2007, so f(-1) = 2007–2008+2009–2010+...+4013–4014. Pairing up the 2008 many summands, each pair having sum -1, the total is -1004.
Or, one can plug in -1 into the first many functions, oberving the sequence of outputs -1, 3, -2, 6, -3, 9, -4, ..., and see the simple pattern -1, -2, -3, ... in the 1st, 3rd, 5th, ... terms. When n is odd, apparently the n’th term is - , so the 2007’th term is -1004.
17. A right triangle is inscribed in a semicircle of radius . Its hypotenuse is the semicircle’s diameter, and its area is half that of the semicircle. Find the perimeter of the triangle, simplified.
Answer: π
Solution: Letting x,y be the lengths of the legs, we have that x2+y2 = 4r2 and xy = πr2. So, 2xy = πr2. So, (x+y)2 = x2+y2+2xy = 4r2+πr2
= (4+π)r2, from which x+y = r. Thus the perimeter is
x+y+2r = r + 2r = ( +2) r = ( +2)( –2) = 4+π – 4 = π.
18. Let S be a solid 2 by 2 square in the plane, and let T be the set of points at distance exactly 1 from the nearest point of S. Find the length of T.
Answer: 8+2π, or 8+2π units
Solution: T consists of four segments each of length 2, and four quarter-circles each of radius 1, for a combined length of 4(2)+4(2π/4) = 8+2π.
19. An ordinary fair die is rolled four times in succession, the numbers appearing being a,b,c,d in that order, where a,b,c,d Î {1,2,3,4,5,6} as usual. When the points (a,b) and (c,d) are plotted in the x,y-plane, given that (a,b) ≠ (c,d), find the probability that the line joining (a,b) and (c,d) has slope zero.
Answer:
Solution: After the first two rolls produce some plotted point (a,b), there are 35 equally likely outcomes for (c,d) [since we must disallow that
(c,d) = (a,b)]. Of these 35 outcomes, exactly five of them are such that
b = d (which is what it takes for the line joining (a,b) & (c,d) to have slope zero), so the probability is = .
20. A polynomial function f(x) of degree 3 has a local maximum at x = -2 and a local minimum at x = 2. Given that the slope of the tangent line to y = f(x) at x = 0 is -12, find f(1) – f(0).
Answer: -11
Solution: [Requires calculus.] f´(x) must be a polynomial of degree 2, and have ±2 as roots, so it factors as f´(x) = k(x–2)(x+2) for some constant k. From the slope information, -12 = f´(0) = k(–2)(2), so k = 3. Therefore
f´(x) = 3(x–2)(x+2) = 3x2–12. So, f(x) must be x3–12x+C for some constant C. So, f(1)–f(0) = (1–12+C) – (0–0+C) = -11.